27 - Light and Matter · moustache, in units of kg. 36 According to folklore, every time you take a breath, you are inhaling some of the atoms exhaled in Caesar’s last words. Is

from rest with an acceleration of 10 m/s2. This is really only a lowerlimit; if there really was a hole that deep, the fall would actuallytake a longer time than the one you calculate, both because thereis air friction and because gravity gets weaker as you get deeper (atthe center of the earth, g is zero, because the earth is pulling youequally in every direction at once).

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22 How many cubic inches are there in a cubic foot? The answeris not 12.

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23 Assume a dog’s brain is twice as great in diameter as a cat’s,but each animal’s brain cells are the same size and their brains arethe same shape. In addition to being a far better companion andmuch nicer to come home to, how many times more brain cells doesa dog have than a cat? The answer is not 2.

24 The population density of Los Angeles is about 4000 people/km2.That of San Francisco is about 6000 people/km2. How many timesfarther away is the average person’s nearest neighbor in LA than inSan Francisco? The answer is not 1.5.

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25 A hunting dog’s nose has about 10 square inches of activesurface. How is this possible, since the dog’s nose is only about 1 in× 1 in × 1 in = 1 in3? After all, 10 is greater than 1, so how can itfit?

26 Estimate the number of blades of grass on a football field.

27 In a computer memory chip, each bit of information (a 0 ora 1) is stored in a single tiny circuit etched onto the surface of asilicon chip. The circuits cover the surface of the chip like lots in ahousing development. A typical chip stores 64 Mb (megabytes) ofdata, where a byte is 8 bits. Estimate (a) the area of each circuit,and (b) its linear size.

28 Suppose someone built a gigantic apartment building, mea-suring 10 km × 10 km at the base. Estimate how tall the buildingwould have to be to have space in it for the entire world’s populationto live.

29 A hamburger chain advertises that it has sold 10 billionBongo Burgers. Estimate the total mass of feed required to raisethe cows used to make the burgers.

30 Estimate the volume of a human body, in cm3.

31 How many cm2 is 1 mm2? . Solution, p. 1038

32 Compare the light-gathering powers of a 3-cm-diameter tele-scope and a 30-cm telescope. . Solution, p. 1038

33 One step on the Richter scale corresponds to a factor of 100in terms of the energy absorbed by something on the surface of theEarth, e.g., a house. For instance, a 9.3-magnitude quake would

Problems 51

Albert Einstein, and his mous-tache, problem 35.

Problem 40.

release 100 times more energy than an 8.3. The energy spreads outfrom the epicenter as a wave, and for the sake of this problem we’llassume we’re dealing with seismic waves that spread out in threedimensions, so that we can visualize them as hemispheres spreadingout under the surface of the earth. If a certain 7.6-magnitude earth-quake and a certain 5.6-magnitude earthquake produce the sameamount of vibration where I live, compare the distances from myhouse to the two epicenters. . Solution, p. 1038

34 In Europe, a piece of paper of the standard size, called A4,is a little narrower and taller than its American counterpart. Theratio of the height to the width is the square root of 2, and this hassome useful properties. For instance, if you cut an A4 sheet from leftto right, you get two smaller sheets that have the same proportions.You can even buy sheets of this smaller size, and they’re called A5.There is a whole series of sizes related in this way, all with the sameproportions. (a) Compare an A5 sheet to an A4 in terms of area andlinear size. (b) The series of paper sizes starts from an A0 sheet,which has an area of one square meter. Suppose we had a seriesof boxes defined in a similar way: the B0 box has a volume of onecubic meter, two B1 boxes fit exactly inside an B0 box, and so on.What would be the dimensions of a B0 box?

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35 Estimate the mass of one of the hairs in Albert Einstein’smoustache, in units of kg.

36 According to folklore, every time you take a breath, you areinhaling some of the atoms exhaled in Caesar’s last words. Is thistrue? If so, how many?

37 The Earth’s surface is about 70% water. Mars’s diameter isabout half the Earth’s, but it has no surface water. Compare theland areas of the two planets.

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38 The traditional Martini glass is shaped like a cone withthe point at the bottom. Suppose you make a Martini by pouringvermouth into the glass to a depth of 3 cm, and then adding ginto bring the depth to 6 cm. What are the proportions of gin andvermouth? . Solution, p. 1038

39 The central portion of a CD is taken up by the hole and somesurrounding clear plastic, and this area is unavailable for storingdata. The radius of the central circle is about 35% of the outerradius of the data-storing area. What percentage of the CD’s areais therefore lost?

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40 The one-liter cube in the photo has been marked off intosmaller cubes, with linear dimensions one tenth those of the bigone. What is the volume of each of the small cubes?

. Solution, p. 1038

52 Chapter 0 Introduction and Review

Problem 42.

Problem 43.

41 Estimate the number of man-hours required for building theGreat Wall of China. . Solution, p. 1038

42 (a) Using the microscope photo in the figure, estimate themass of a one cell of the E. coli bacterium, which is one of themost common ones in the human intestine. Note the scale at thelower right corner, which is 1 µm. Each of the tubular objects inthe column is one cell. (b) The feces in the human intestine aremostly bacteria (some dead, some alive), of which E. coli is a largeand typical component. Estimate the number of bacteria in yourintestines, and compare with the number of human cells in yourbody, which is believed to be roughly on the order of 1013. (c)Interpreting your result from part b, what does this tell you aboutthe size of a typical human cell compared to the size of a typicalbacterial cell?

43 The figure shows a practical, simple experiment for determin-ing g to high precision. Two steel balls are suspended from electro-magnets, and are released simultaneously when the electric currentis shut off. They fall through unequal heights ∆x1 and ∆x2. Acomputer records the sounds through a microphone as first one balland then the other strikes the floor. From this recording, we canaccurately determine the quantity T defined as T = ∆t2 −∆t1, i.e.,the time lag between the first and second impacts. Note that sincethe balls do not make any sound when they are released, we haveno way of measuring the individual times ∆t2 and ∆t1.(a) Find an equation for g in terms of the measured quantities T ,∆x1 and ∆x2.

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(b) Check the units of your equation.(c) Check that your equation gives the correct result in the casewhere ∆x1 is very close to zero. However, is this case realistic?(d) What happens when ∆x1 = ∆x2? Discuss this both mathemat-ically and physically.

44 Estimate the number of jellybeans in figure o on p. 44.. Solution, p. 1039

45 Let the function x be defined by x(t) = Aebt, where t hasunits of seconds and x has units of meters. (For b < 0, this could bea fairly accurate model of the motion of a bullet shot into a tank ofoil.) Show that the Taylor series of this function makes sense if andonly if A and b have certain units.

46 A 2002 paper by Steegmann et al. uses data from modernhuman groups like the Inuit to argue that Neanderthals in Ice AgeEurope had to eat up “to 4,480 kcal per day to support strenuouswinter foraging and cold resistance costs.” What’s wrong here?

Key to symbols:easy typical challenging difficult very difficult

Problems 53

√An answer check is available at www.lightandmatter.com.

54 Chapter 0 Introduction and Review

The universe has been recyclingits contents ever since the BigBang, 13.7 billion years ago.

Chapter 1

Conservation of Mass

It took just a moment for that head to fall, but a hundred yearsmight not produce another like it.

Joseph-Louis Lagrange, referring to the execution of Lavoisieron May 8, 1794

1.1 MassChange is impossible, claimed the ancient Greek philosopher Par-menides. His work was nonscientific, since he didn’t state his ideasin a form that would allow them to be tested experimentally, butmodern science nevertheless has a strong Parmenidean flavor. Hismain argument that change is an illusion was that something can’tbe turned into nothing, and likewise if you have nothing, you can’tturn it into something. To make this into a scientific theory, we haveto decide on a way to measure what “something” is, and we can then

55

a / Portrait of Monsieur Lavoisierand His Wife, by Jacques-LouisDavid, 1788. Lavoisier inventedthe concept of conservation ofmass. The husband is depictedwith his scientific apparatus,while in the background on theleft is the portfolio belongingto Madame Lavoisier, who isthought to have been a student ofDavid’s.

b / A measurement of gravi-tational mass: the sphere hasa gravitational mass of fivekilograms.

check by measurements whether the total amount of “something” inthe universe really stays constant. How much “something” is therein a rock? Does a sunbeam count as “something?” Does heat count?Motion? Thoughts and feelings?

If you look at the table of contents of this book, you’ll see thatthe first four chapters have the word “conservation” in them. Inphysics, a conservation law is a statement that the total amount ofa certain physical quantity always stays the same. This chapter isabout conservation of mass. The metric system is designed arounda unit of distance, the meter, a unit of mass, the kilogram, and atime unit, the second. Numerical measurement of distance and timeprobably date back almost as far into prehistory as counting money,but mass is a more modern concept. Until scientists figured out thatmass was conserved, it wasn’t obvious that there could be a single,consistent way of measuring an amount of matter, hence jiggers ofwhiskey and cords of wood. You may wonder why conservation ofmass wasn’t discovered until relatively modern times, but it wasn’tobvious, for example, that gases had mass, and that the apparentloss of mass when wood was burned was exactly matched by themass of the escaping gases.

Once scientists were on the track of the conservation of massconcept, they began looking for a way to define mass in terms of adefinite measuring procedure. If they tried such a procedure, and theresult was that it led to nonconservation of mass, then they wouldthrow it out and try a different procedure. For instance, we mightbe tempted to define mass using kitchen measuring cups, i.e., as ameasure of volume. Mass would then be perfectly conserved for aprocess like mixing marbles with peanut butter, but there would beprocesses like freezing water that led to a net increase in mass, andothers like soaking up water with a sponge that caused a decrease.If, with the benefit of hindsight, it seems like the measuring cupdefinition was just plain silly, then here’s a more subtle example ofa wrong definition of mass. Suppose we define it using a bathroomscale, or a more precise device such as a postal scale that works onthe same principle of using gravity to compress or twist a spring.The trouble is that gravity is not equally strong all over the surfaceof the earth, so for instance there would be nonconservation of masswhen you brought an object up to the top of a mountain, wheregravity is a little weaker.

Although some of the obvious possibilities have problems, theredo turn out to be at least two approaches to defining mass that leadto its being a conserved quantity, so we consider these definitionsto be “right” in the pragmatic sense that what’s correct is what’suseful.

One definition that works is to use balances, but compensatefor the local strength of gravity. This is the method that is used

56 Chapter 1 Conservation of Mass

c / A measurement of inertialmass: the wagon recoils withthe same velocity in experiments1 and 2, establishing that theinertial mass of the cement blockis four kilograms.

d / The time for one cycle ofvibration is related to the object’sinertial mass.

e / Astronaut Tamara Jerniganmeasures her inertial massaboard the Space Shuttle.

by scientists who actually specialize in ultraprecise measurements.A standard kilogram, in the form of a platinum-iridium cylinder,is kept in a special shrine in Paris. Copies are made that balanceagainst the standard kilogram in Parisian gravity, and they are thentransported to laboratories in other parts of the world, where theyare compared with other masses in the local gravity. The quantitydefined in this way is called gravitational mass.

A second and completely different approach is to measure howhard it is to change an object’s state of motion. This tells us its in-ertial mass. For example, I’d be more willing to stand in the way ofan oncoming poodle than in the path of a freight train, because mybody will have a harder time convincing the freight train to stop.This is a dictionary-style conceptual definition, but in physics weneed to back up a conceptual definition with an operational defini-tion, which is one that spells out the operations required in orderto measure the quantity being defined. We can operationalize ourdefinition of inertial mass by throwing a standard kilogram at an ob-ject at a speed of 1 m/s (one meter per second) and measuring therecoiling object’s velocity. Suppose we want to measure the massof a particular block of cement. We put the block in a toy wagonon the sidewalk, and throw a standard kilogram at it. Suppose thestandard kilogram hits the wagon, and then drops straight downto the sidewalk, having lost all its velocity, and the wagon and theblock inside recoil at a velocity of 0.23 m/s. We then repeat theexperiment with the block replaced by various numbers of standardkilograms, and find that we can reproduce the recoil velocity of 0.23m/s with four standard kilograms in the wagon. We have deter-mined the mass of the block to be four kilograms.1 Although thisdefinition of inertial mass has an appealing conceptual simplicity, itis obviously not very practical, at least in this crude form. Never-theless, this method of collision is very much like the methods usedfor measuring the masses of subatomic particles, which, after all,can’t be put on little postal scales!

Astronauts spending long periods of time in space need to mon-itor their loss of bone and muscle mass, and here as well, it’s im-possible to measure gravitational mass. Since they don’t want tohave standard kilograms thrown at them, they use a slightly differ-ent technique (figures d and e). They strap themselves to a chairwhich is attached to a large spring, and measure the time it takesfor one cycle of vibration.

1You might think intuitively that the recoil velocity should be exactly onefourth of a meter per second, and you’d be right except that the wagon has somemass as well. Our present approach, however, only requires that we give a wayto test for equality of masses. To predict the recoil velocity from scratch, we’dneed to use conservation of momentum, which is discussed in a later chapter.

Section 1.1 Mass 57

f / Example 1.

1.1.1 Problem-solving techniques

How do we use a conservation law, such as conservation of mass,to solve problems? There are two basic techniques.

As an analogy, consider conservation of money, which makes itillegal for you to create dollar bills using your own inkjet printer.(Most people don’t intentionally destroy their dollar bills, either!)Suppose the police notice that a particular store doesn’t seem tohave any customers, but the owner wears lots of gold jewelry anddrives a BMW. They suspect that the store is a front for some kindof crime, perhaps counterfeiting. With intensive surveillance, thereare two basic approaches they could use in their investigation. Onemethod would be to have undercover agents try to find out howmuch money goes in the door, and how much money comes backout at the end of the day, perhaps by arranging through some trickto get access to the owner’s briefcase in the morning and evening. Ifthe amount of money that comes out every day is greater than theamount that went in, and if they’re convinced there is no safe on thepremises holding a large reservoir of money, then the owner mustbe counterfeiting. This inflow-equals-outflow technique is useful ifwe are sure that there is a region of space within which there is nosupply of mass that is being built up or depleted.

A stream of water example 1If you watch water flowing out of the end of a hose, you’ll seethat the stream of water is fatter near the mouth of the hose, andskinnier lower down. This is because the water speeds up as itfalls. If the cross-sectional area of the stream was equal all alongits length, then the rate of flow (kilograms per second) through alower cross-section would be greater than the rate of flow througha cross-section higher up. Since the flow is steady, the amountof water between the two cross-sections stays constant. Conser-vation of mass therefore requires that the cross-sectional area ofthe stream shrink in inverse proportion to the increasing speed ofthe falling water.

self-check ASuppose the you point the hose straight up, so that the water is risingrather than falling. What happens as the velocity gets smaller? Whathappens when the velocity becomes zero? . Answer, p. 1058

How can we apply a conservation law, such as conservation ofmass, in a situation where mass might be stored up somewhere? Touse a crime analogy again, a prison could contain a certain numberof prisoners, who are not allowed to flow in or out at will. In physics,this is known as a closed system. A guard might notice that a certainprisoner’s cell is empty, but that doesn’t mean he’s escaped. Hecould be sick in the infirmary, or hard at work in the shop earningcigarette money. What prisons actually do is to count all theirprisoners every day, and make sure today’s total is the same as


yesterday’s. One way of stating a conservation law is that for aclosed system, the total amount of stuff (mass, in this chapter) staysconstant.

Lavoisier and chemical reactions in a closed system example 2The French chemist Antoine-Laurent Lavoisier is considered theinventor of the concept of conservation of mass. Before Lavoisier,chemists had never systematically weighed their chemicals toquantify the amount of each substance that was undergoing re-actions. They also didn’t completely understand that gases werejust another state of matter, and hadn’t tried performing reactionsin sealed chambers to determine whether gases were being con-sumed from or released into the air. For this they had at leastone practical excuse, which is that if you perform a gas-releasingreaction in a sealed chamber with no room for expansion, you getan explosion! Lavoisier invented a balance that was capable ofmeasuring milligram masses, and figured out how to do reactionsin an upside-down bowl in a basin of water, so that the gasescould expand by pushing out some of the water. In a crucial ex-periment, Lavoisier heated a red mercury compound, which wewould now describe as mercury oxide (HgO), in such a sealedchamber. A gas was produced (Lavoisier later named it “oxy-gen”), driving out some of the water, and the red compound wastransformed into silvery liquid mercury metal. The crucial pointwas that the total mass of the entire apparatus was exactly thesame before and after the reaction. Based on many observationsof this type, Lavoisier proposed a general law of nature, that massis always conserved. (In earlier experiments, in which closed sys-tems were not used, chemists had become convinced that therewas a mysterious substance, phlogiston, involved in combustionand oxidation reactions, and that phlogiston’s mass could be pos-itive, negative, or zero depending on the situation!)

1.1.2 Delta notation

A convenient notation used throughout physics is ∆, the up-percase Greek letter delta, which indicates “change in” or “afterminus before.” For example, if b represents how much money youhave in the bank, then a deposit of $100 could be represented as∆b = $100. That is, the change in your balance was $100, or thebalance after the transaction minus the balance before the transac-tion equals $100. A withdrawal would be indicated by ∆b < 0. Werepresent “before” and “after” using the subscripts i (initial) andf (final), e.g., ∆b = bf − bi. Often the delta notation allows moreprecision than English words. For instance, “time” can be used tomean a point in time (“now’s the time”), t, or it could mean a periodof time (“the whole time, he had spit on his chin”), ∆t.

This notation is particularly convenient for discussing conservedquantities. The law of conservation of mass can be stated simply as

Section 1.1 Mass 59

a / The two pendulum bobsare constructed with equal grav-itational masses. If their inertialmasses are also equal, then eachpendulum should take exactly thesame amount of time per swing.

b / If the cylinders have slightlyunequal ratios of inertial to grav-itational mass, their trajectorieswill be a little different.

c / A simplified drawing of anEotvos-style experiment. Ifthe two masses, made out oftwo different substances, haveslightly different ratios of inertialto gravitational mass, then theapparatus will twist slightly as theearth spins.

∆m = 0, where m is the total mass of any closed system.

self-check BIf x represents the location of an object moving in one dimension, thenhow would positive and negative signs of ∆x be interpreted? .

Answer, p. 1058

Discussion Questions

A If an object had a straight-line x − t graph with ∆x = 0 and ∆t 6= 0,what would be true about its velocity? What would this look like on agraph? What about ∆t = 0 and ∆x 6= 0?

1.2 Equivalence of gravitational and inertial massWe find experimentally that both gravitational and inertial massare conserved to a high degree of precision for a great number ofprocesses, including chemical reactions, melting, boiling, soaking upwater with a sponge, and rotting of meat and vegetables. Now it’slogically possible that both gravitational and inertial mass are con-served, but that there is no particular relationship between them, inwhich case we would say that they are separately conserved. On theother hand, the two conservation laws may be redundant, like havingone law against murder and another law against killing people!

Here’s an experiment that gets at the issue: stand up now anddrop a coin and one of your shoes side by side. I used a 400-gramshoe and a 2-gram penny, and they hit the floor at the same timeas far as I could tell by eye. This is an interesting result, but aphysicist and an ordinary person will find it interesting for differentreasons.

The layperson is surprised, since it would seem logical thatheaver objects would always fall faster than light ones. However,it’s fairly easy to prove that if air friction is negligible, any two ob-jects made of the same substance must have identical motion whenthey fall. For instance, a 2-kg copper mass must exhibit the samefalling motion as a 1-kg copper mass, because nothing would bechanged by physically joining together two 1-kg copper masses tomake a single 2-kg copper mass. Suppose, for example, that theyare joined with a dab of glue; the glue isn’t under any strain, be-cause the two masses are doing the same thing side by side. Sincethe glue isn’t really doing anything, it makes no difference whetherthe masses fall separately or side by side.2

What a physicist finds remarkable about the shoe-and-penny ex-periment is that it came out the way it did even though the shoeand the penny are made of different substances. There is absolutelyno theoretical reason why this should be true. We could say that it

2The argument only fails for objects light enough to be affected appreciablyby air friction: a bunch of feathers falls differently if you wad them up becausethe pattern of air flow is altered by putting them together.


d / A more realistic drawingof Braginskii and Panov’s ex-periment. The whole thing wasencased in a tall vacuum tube,which was placed in a sealedbasement whose temperaturewas controlled to within 0.02◦C.The total mass of the platinumand aluminum test masses,plus the tungsten wire and thebalance arms, was only 4.4 g.To detect tiny motions, a laserbeam was bounced off of a mirrorattached to the wire. There wasso little friction that the balancewould have taken on the order ofseveral years to calm down com-pletely after being put in place;to stop these vibrations, staticelectrical forces were appliedthrough the two circular plates toprovide very gentle twists on theellipsoidal mass between them.After Braginskii and Panov.

happens because the greater gravitational mass of the shoe is exactlycounteracted by its greater inertial mass, which makes it harder forgravity to get it moving, but that just leaves us wondering why in-ertial mass and gravitational mass are always in proportion to eachother. It’s possible that they are only approximately equivalent.Most of the mass of ordinary matter comes from neutrons and pro-tons, and we could imagine, for instance, that neutrons and protonsdo not have exactly the same ratio of gravitational to inertial mass.This would show up as a different ratio of gravitational to inertialmass for substances containing different proportions of neutrons andprotons.

Galileo did the first numerical experiments on this issue in theseventeenth century by rolling balls down inclined planes, althoughhe didn’t think about his results in these terms. A fairly easy way toimprove on Galileo’s accuracy is to use pendulums with bobs madeof different materials. Suppose, for example, that we construct analuminum bob and a brass bob, and use a double-pan balance toverify to good precision that their gravitational masses are equal. Ifwe then measure the time required for each pendulum to performa hundred cycles, we can check whether the results are the same.If their inertial masses are unequal, then the one with a smallerinertial mass will go through each cycle faster, since gravity hasan easier time accelerating and decelerating it. With this type ofexperiment, one can easily verify that gravitational and inertial massare proportional to each other to an accuracy of 10−3 or 10−4.

In 1889, the Hungarian physicist Roland Eotvos used a slightlydifferent approach to verify the equivalence of gravitational and in-ertial mass for various substances to an accuracy of about 10−8, andthe best such experiment, figure d, improved on even this phenome-nal accuracy, bringing it to the 10−12 level.3 In all the experimentsdescribed so far, the two objects move along similar trajectories:straight lines in the penny-and-shoe and inclined plane experiments,and circular arcs in the pendulum version. The Eotvos-style exper-iment looks for differences in the objects’ trajectories. The conceptcan be understood by imagining the following simplified version.Suppose, as in figure b, we roll a brass cylinder off of a tabletopand measure where it hits the floor, and then do the same with analuminum cylinder, making sure that both of them go over the edgewith precisely the same velocity. An object with zero gravitationalmass would fly off straight and hit the wall, while an object withzero inertial mass would make a sudden 90-degree turn and dropstraight to the floor. If the aluminum and brass cylinders have or-dinary, but slightly unequal, ratios of gravitational to inertial mass,then they will follow trajectories that are just slightly different. Inother words, if inertial and gravitational mass are not exactly pro-portional to each other for all substances, then objects made of

3V.B. Braginskii and V.I. Panov, Soviet Physics JETP 34, 463 (1972).

Section 1.2 Equivalence of gravitational and inertial mass 61

a / Galileo Galilei (1564-1642).

different substances will have different trajectories in the presenceof gravity.

A simplified drawing of a practical, high-precision experimentis shown in figure c. Two objects made of different substances arebalanced on the ends of a bar, which is suspended at the center froma thin fiber. The whole apparatus moves through space on a com-plicated, looping trajectory arising from the rotation of the earthsuperimposed on the earth’s orbital motion around the sun. Boththe earth’s gravity and the sun’s gravity act on the two objects. Iftheir inertial masses are not exactly in proportion to their gravi-tational masses, then they will follow slightly different trajectoriesthrough space, which will result in a very slight twisting of the fiberbetween the daytime, when the sun’s gravity is pulling upward, andthe night, when the sun’s gravity is downward. Figure d shows amore realistic picture of the apparatus.

This type of experiment, in which one expects a null result, isa tough way to make a career as a scientist. If your measurementcomes out as expected, but with better accuracy than other peoplehad previously achieved, your result is publishable, but won’t beconsidered earthshattering. On the other hand, if you build themost sensitive experiment ever, and the result comes out contraryto expectations, you’re in a scary situation. You could be right, andearn a place in history, but if the result turns out to be due to adefect in your experiment, then you’ve made a fool of yourself.

1.3 Galilean relativity

I defined inertial mass conceptually as a measure of how hard itis to change an object’s state of motion, the implication being that ifyou don’t interfere, the object’s motion won’t change. Most people,however, believe that objects in motion have a natural tendency toslow down. Suppose I push my refrigerator to the west for a while at0.1 m/s, and then stop pushing. The average person would say fridgejust naturally stopped moving, but let’s imagine how someone inChina would describe the fridge experiment carried out in my househere in California. Due to the rotation of the earth, California ismoving to the east at about 400 m/s. A point in China at the samelatitude has the same speed, but since China is on the other sideof the planet, China’s east is my west. (If you’re finding the three-dimensional visualization difficult, just think of China and Californiaas two freight trains that go past each other, each traveling at 400m/s.) If I insist on thinking of my dirt as being stationary, thenChina and its dirt are moving at 800 m/s to my west. From China’spoint of view, however, it’s California that is moving 800 m/s inthe opposite direction (my east). When I’m pushing the fridge tothe west at 0.1 m/s, the observer in China describes its speed as799.9 m/s. Once I stop pushing, the fridge speeds back up to 800


b / The earth spins. Peoplein Shanghai say they’re at restand people in Los Angeles aremoving. Angelenos say the sameabout the Shanghainese.

m/s. From my point of view, the fridge “naturally” slowed downwhen I stopped pushing, but according to the observer in China, it“naturally” sped up!

What’s really happening here is that there’s a tendency, dueto friction, for the fridge to stop moving relative to the floor. Ingeneral, only relative motion has physical significance in physics, notabsolute motion. It’s not even possible to define absolute motion,since there is no special reference point in the universe that everyonecan agree is at rest. Of course if we want to measure motion, wedo have to pick some arbitrary reference point which we will sayis standing still, and we can then define x, y, and z coordinatesextending out from that point, which we can define as having x = 0,y = 0, z = 0. Setting up such a system is known as choosing aframe of reference. The local dirt is a natural frame of reference fordescribing a game of basketball, but if the game was taking place onthe deck of a moving ocean liner, we would probably pick a frame ofreference in which the deck was at rest, and the land was moving.

Galileo was the first scientist to reason along these lines, andwe now use the term Galilean relativity to refer to a somewhatmodernized version of his principle. Roughly speaking, the principleof Galilean relativity states that the same laws of physics apply inany frame of reference that is moving in a straight line at constantspeed. We need to refine this statement, however, since it is notnecessarily obvious which frames of reference are going in a straightline at constant speed. A person in a pickup truck pulling awayfrom a stoplight could admit that the car’s velocity is changing,or she could insist that the truck is at rest, and the meter on thedashboard is going up because the asphalt picked that moment tostart moving faster and faster backward! Frames of reference arenot all created equal, however, and the accelerating truck’s frameof reference is not as good as the asphalt’s. We can tell, becausea bowling ball in the back of the truck, as in figure c, appears tobehave strangely in the driver’s frame of reference: in her rear-view mirror, she sees the ball, initially at rest, start moving fasterand faster toward the back of the truck. This goofy behavior isevidence that there is something wrong with her frame of reference.A person on the sidewalk, however, sees the ball as standing still. Inthe sidewalk’s frame of reference, the truck pulls away from the ball,and this makes sense, because the truck is burning gas and using upenergy to change its state of motion.

We therefore define an inertial frame of reference as one in whichwe never see objects change their state of motion without any appar-ent reason. The sidewalk is a pretty good inertial frame, and a carmoving relative to the sidewalk at constant speed in a straight linedefines a pretty good inertial frame, but a car that is acceleratingor turning is not a inertial frame.

Section 1.3 Galilean relativity 63

c / Left: In a frame of referencethat speeds up with the truck, thebowling ball appears to changeits state of motion for no reason.Right: In an inertial frame of ref-erence, which the surface of theearth approximately is, the bowl-ing ball stands still, which makessense because there is nothingthat would cause it to change itsstate of motion.

The principle of Galilean relativity states that inertial framesexist, and that the same laws of physics apply in all inertial framesof reference, regardless of one frame’s straight-line, constant-speedmotion relative to another.4

Another way of putting it is that all inertial frames are createdequal. We can say whether one inertial frame is in motion or at restrelative to another, but there is no privileged “rest frame.” Thereis no experiment that comes out any different in laboratories indifferent inertial frames, so there is no experiment that could tell uswhich inertial frame is really, truly at rest.

The speed of sound example 3. The speed of sound in air is only 340 m/s, so unless you liveat a near-polar latitude, you’re moving at greater than the speedof sound right now due to the Earth’s rotation. In that case, whydon’t we experience exciting phenomena like sonic booms all thetime? . It might seem as though you’re unprepared to deal withthis question right now, since the only law of physics you knowis conservation of mass, and conservation of mass doesn’t tellyou anything obviously useful about the speed of sound or sonicbooms. Galilean relativity, however, is a blanket statement aboutall the laws of physics, so in a situation like this, it may let youpredict the results of the laws of physics without actually knowingwhat all the laws are! If the laws of physics predict a certain valuefor the speed of sound, then they had better predict the speed

4The principle of Galilean relativity is extended on page 195.


d / Foucault demonstrateshis pendulum to an audience at alecture in 1851.

e / Galileo’s trial.

of the sound relative to the air, not their speed relative to somespecial “rest frame.” Since the air is moving along with the rotationof the earth, we don’t detect any special phenomena. To get asonic boom, the source of the sound would have to be movingrelative to the air.

The Foucault pendulum example 4Note that in the example of the bowling ball in the truck, I didn’tclaim the sidewalk was exactly a Galilean frame of reference.This is because the sidewalk is moving in a circle due to the rota-tion of the Earth, and is therefore changing the direction of its mo-tion continuously on a 24-hour cycle. However, the curve of themotion is so gentle that under ordinary conditions we don’t noticethat the local dirt’s frame of reference isn’t quite inertial. The firstdemonstration of the noninertial nature of the earth-fixed frame ofreference was by Foucault using a very massive pendulum (fig-ure d) whose oscillations would persist for many hours without be-coming imperceptible. Although Foucault did his demonstration inParis, it’s easier to imagine what would happen at the north pole:the pendulum would keep swinging in the same plane, but theearth would spin underneath it once every 24 hours. To someonestanding in the snow, it would appear that the pendulum’s planeof motion was twisting. The effect at latitudes less than 90 de-grees turns out to be slower, but otherwise similar. The Foucaultpendulum was the first definitive experimental proof that the earthreally did spin on its axis, although scientists had been convincedof its rotation for a century based on more indirect evidence aboutthe structure of the solar system.

Although popular belief has Galileo being prosecuted by theCatholic Church for saying the earth rotated on its axis and also or-bited the sun, Foucault’s pendulum was still centuries in the future,so Galileo had no hard proof; Galileo’s insights into relative versusabsolute motion simply made it more plausible that the world couldbe spinning without producing dramatic effects, but didn’t disprovethe contrary hypothesis that the sun, moon, and stars went aroundthe earth every 24 hours. Furthermore, the Church was much moreliberal and enlightened than most people believe. It didn’t (and stilldoesn’t) require a literal interpretation of the Bible, and one of theChurch officials involved in the Galileo affair wrote that “the Bibletells us how to go to heaven, not how the heavens go.” In otherwords, religion and science should be separate. The actual reasonGalileo got in trouble is shrouded in mystery, since Italy in the age ofthe Medicis was a secretive place where unscrupulous people mightsettle a score with poison or a false accusation of heresy. What is cer-tain is that Galileo’s satirical style of scientific writing made manyenemies among the powerful Jesuit scholars who were his intellec-tual opponents — he compared one to a snake that doesn’t knowits own back is broken. It’s also possible that the Church was far


f / Discussion question A.

g / Discussion question B.

h / Discussion question D.

less upset by his astronomical work than by his support for atomism(discussed further in the next section). Some theologians perceivedatomism as contradicting transubstantiation, the Church’s doctrinethat the holy bread and wine were literally transformed into theflesh and blood of Christ by the priest’s blessing.

self-check CWhat is incorrect about the following supposed counterexamples to theprinciple of inertia?

(1) When astronauts blast off in a rocket, their huge velocity does causea physical effect on their bodies — they get pressed back into theirseats, the flesh on their faces gets distorted, and they have a hard timelifting their arms.

(2) When you’re driving in a convertible with the top down, the wind inyour face is an observable physical effect of your absolute motion. .

Answer, p. 1058

. Solved problem: a bug on a wheel page 71, problem 12


A A passenger on a cruise ship finds, while the ship is docked, thathe can leap off of the upper deck and just barely make it into the poolon the lower deck. If the ship leaves dock and is cruising rapidly, will thisadrenaline junkie still be able to make it?

B You are a passenger in the open basket hanging under a heliumballoon. The balloon is being carried along by the wind at a constantvelocity. If you are holding a flag in your hand, will the flag wave? If so,which way? [Based on a question from PSSC Physics.]

C Aristotle stated that all objects naturally wanted to come to rest, withthe unspoken implication that “rest” would be interpreted relative to thesurface of the earth. Suppose we could transport Aristotle to the moon,put him in a space suit, and kick him out the door of the spaceship and intothe lunar landscape. What would he expect his fate to be in this situation?If intelligent creatures inhabited the moon, and one of them independentlycame up with the equivalent of Aristotelian physics, what would they thinkabout objects coming to rest?

D Sally is on an amusement park ride which begins with her chair beinghoisted straight up a tower at a constant speed of 60 miles/hour. Despitestern warnings from her father that he’ll take her home the next time shemisbehaves, she decides that as a scientific experiment she really needsto release her corndog over the side as she’s on the way up. She doesnot throw it. She simply sticks it out of the car, lets it go, and watches itagainst the background of the sky, with no trees or buildings as referencepoints. What does the corndog’s motion look like as observed by Sally?Does its speed ever appear to her to be zero? What acceleration doesshe observe it to have: is it ever positive? negative? zero? What wouldher enraged father answer if asked for a similar description of its motionas it appears to him, standing on the ground?


1.3.1 Applications of calculus

Let’s see how this relates to calculus. If an object is moving inone dimension, we can describe its position with a function x(t). Thederivative v = dx/ dt is called the velocity, and the second derivativea = dv/dt = d2x/ dt2 is the acceleration. Galilean relativity tells usthat there is no detectable effect due to an object’s absolute velocity,since in some other frame of reference, the object’s velocity mightbe zero. However, an acceleration does have physical consequences.

i / This Air Force doctor volun-teered to ride a rocket sled as amedical experiment. The obviouseffects on his head and face arenot because of the sled’s speedbut because of its rapid changesin speed: increasing in (ii) and(iii), and decreasing in (v) and (vi).In (iv) his speed is greatest, butbecause his speed is not increas-ing or decreasing very much atthis moment, there is little effecton him. (U.S. Air Force)

Observers in different inertial frames of reference will disagreeon velocities, but agree on accelerations. Let’s keep it simple bycontinuing to work in one dimension. One frame of reference uses acoordinate system x1, and the other we label x2. If the positive x1

and x2 axes point in the same direction, then in general two inertialframes could be related by an equation of the form x2 = x1 + b+ut,where u is the constant velocity of one frame relative to the other,and the constant b tells us how far apart the origins of the twocoordinate systems were at t = 0. The velocities are different in thetwo frames of reference:

dx2

dt=

dx1

dt+ u,

Suppose, for example, frame 1 is defined from the sidewalk, andframe 2 is fixed to a float in a parade that is moving to our left ata velocity u = 1 m/s. A dog that is moving to the right with avelocity v1 = dx1/ dt = 3 m/s in the sidewalk’s frame will appearto be moving at a velocity of v2 = dx2/ dt = dx1/ dt+u = 4 m/s inthe float’s frame.


Self-check D.

For acceleration, however, we have

d2 x2

dt2=

d2 x1

dt2,

since the derivative of the constant u is zero. Thus an accelera-tion, unlike a velocity, can have a definite physical significance to allobservers in all frames of reference. If this wasn’t true, then therewould be no particular reason to define a quantity called accelerationin the first place.

self-check DThe figure shows a bottle of beer sitting on a table in the dining car of atrain. Does the tilting of the surface tell us about the train’s velocity, orits acceleration? What would a person in the train say about the bottle’svelocity? What about a person standing in a field outside and looking inthrough the window? What about the acceleration? . Answer, p.1058

1.4 A preview of some modern physics“Mommy, why do you and Daddy have to go to work?” “To makemoney, sweetie-pie.” “Why do we need money?” “To buy food.”“Why does food cost money?” When small children ask a chainof “why” questions like this, it usually isn’t too long before theirparents end up saying something like, “Because that’s just the wayit is,” or, more honestly, “I don’t know the answer.”

The same happens in physics. We may gradually learn to ex-plain things more and more deeply, but there’s always the possibilitythat a certain observed fact, such as conservation of mass, will neverbe understood on any deeper level. Science, after all, uses limitedmethods to achieve limited goals, so the ultimate reason for all exis-tence will always be the province of religion. There is, however, anappealing explanation for conservation of mass, which is atomism,the theory that matter is made of tiny, unchanging particles. Theatomic hypothesis dates back to ancient Greece, but the first solidevidence to support it didn’t come until around the eighteenth cen-tury, and individual atoms were never detected until about 1900.The atomic theory implies not only conservation of mass, but acouple of other things as well.

First, it implies that the total mass of one particular elementis conserved. For instance, lead and gold are both elements, and ifwe assume that lead atoms can’t be turned into gold atoms, thenthe total mass of lead and the total mass of gold are separatelyconserved. It’s as though there was not just a law against pickpock-eting, but also a law against surreptitiously moving money fromone of the victim’s pockets to the other. It turns out, however, thatalthough chemical reactions never change one type of atom into an-other, transmutation can happen in nuclear reactions, such as the


ones that created most of the elements in your body out of the pri-mordial hydrogen and helium that condensed out of the aftermathof the Big Bang.

Second, atomism implies that mass is quantized, meaning thatonly certain values of mass are possible and the ones in betweencan’t exist. We can have three atoms of gold or four atoms of gold,but not three an a half. Although quantization of mass is a naturalconsequence of any theory in which matter is made up of tiny parti-cles, it was discovered in the twentieth century that other quantities,such as energy, are quantized as well, which had previously not beensuspected.

self-check EIs money quantized? . Answer, p. 1058

If atomism is starting to make conservation of mass seem in-evitable to you, then it may disturb you to know that Einsteindiscovered it isn’t really conserved. If you put a 50-gram iron nailin some water, seal the whole thing up, and let it sit on a fantasti-cally precise balance while the nail rusts, you’ll find that the systemloses about 6 × 10−12 kg of mass by the time the nail has turnedcompletely to rust. This has to do with Einstein’s famous equa-tion E = mc2. Rusting releases heat energy, which then escapesout into the room. Einstein’s equation states that this amount ofheat, E, is equivalent to a certain amount of mass, m. The c inthe c2 is the speed of light, which is a large number, and a largeamount of energy is therefore equivalent to a very small amount ofmass, so you don’t notice nonconservation of mass under ordinaryconditions. What is really conserved is not the mass, m, but themass-plus-energy, E + mc2. The point of this discussion is not toget you to do numerical exercises with E = mc2 (at this point youdon’t even know what units are used to measure energy), but simplyto point out to you the empirical nature of the laws of physics. If apreviously accepted theory is contradicted by an experiment, thenthe theory needs to be changed. This is also a good example ofsomething called the correspondence principle, which is a historicalobservation about how scientific theories change: when a new scien-tific theory replaces an old one, the old theory is always containedwithin the new one as an approximation that works within a certainrestricted range of situations. Conservation of mass is an extremelygood approximation for all chemical reactions, since chemical reac-tions never release or consume enough energy to change the totalmass by a large percentage. Conservation of mass would not havebeen accepted for 110 years as a fundamental principle of physicsif it hadn’t been verified over and over again by a huge number ofaccurate experiments.

This chapter is summarized on page 1075. Notation and terminologyare tabulated on pages 1070-1071.

Section 1.4 A preview of some modern physics 69

Problem 6.

Problem 8.

ProblemsThe symbols

√, , etc. are explained on page 72.

1 Thermometers normally use either mercury or alcohol as theirworking fluid. If the level of the fluid rises or falls, does this violateconservation of mass?

2 The ratios of the masses of different types of atoms weredetermined a century before anyone knew any actual atomic massesin units of kg. One finds, for example, that when ordinary tablesalt, NaCl, is melted, the chlorine atoms bubble off as a gas, leavingliquid sodium metal. Suppose the chlorine escapes, so that its masscannot be directly determined by weighing. Experiments show thatwhen 1.00000 kg of NaCl is treated in this way, the mass of theremaining sodium metal is 0.39337 kg. Based on this information,determine the ratio of the mass of a chlorine atom to that of asodium atom.

√

3 An atom of the most common naturally occurring uraniumisotope breaks up spontaneously into a thorium atom plus a heliumatom. The masses are as follows:

uranium 3.95292849× 10−25 kgthorium 3.88638748× 10−25 kghelium 6.646481× 10−27 kg

Each of these experimentally determined masses is uncertain in itslast decimal place. Is mass conserved in this process to within theaccuracy of the experimental data? How would you interpret this?

4 If two spherical water droplets of radius b combine to make asingle droplet, what is its radius? (Assume that water has constantdensity.)

5 Make up an experiment that would test whether mass is con-served in an animal’s metabolic processes.

6 The figure shows a hydraulic jack. What is the relationshipbetween the distance traveled by the plunger and the distance trav-eled by the object being lifted, in terms of the cross-sectional areas?

7 In an example in this chapter, I argued that a stream of watermust change its cross-sectional area as it rises or falls. Suppose thatthe stream of water is confined to a constant-diameter pipe. Whichassumption breaks down in this situation?

8 A river with a certain width and depth splits into two parts,each of which has the same width and depth as the original river.What can you say about the speed of the current after the split?

9 The diagram shows a cross-section of a wind tunnel of the


Problem 11.

Problem 12.

kind used, for example, to test designs of airplanes. Under normalconditions of use, the density of the air remains nearly constantthroughout the whole wind tunnel. How can the speed of the airbe controlled and calculated? (Diagram by NASA, Glenn ResearchCenter.)

10 A water wave is in a tank that extends horizontally fromx = 0 to x = a, and from z = 0 to z = b. We assume for simplicitythat at a certain moment in time the height y of the water’s surfaceonly depends on x, not z, so that we can effectively ignore the zcoordinate. Under these assumptions, the total volume of the waterin the tank is

V = b

∫ a

0y(x) dx.

Since the density of the water is essentially constant, conservationof mass requires that V is always the same. When the water is calm,we have y = h, where h = V/ab. If two different wave patterns moveinto each other, we might imagine that they would add in the sensethat ytotal−h = (y1−h) + (y2−h). Show that this type of additionis consistent with conservation of mass.

11 The figure shows the position of a falling ball at equal timeintervals, depicted in a certain frame of reference. On a similar grid,show how the ball’s motion would appear in a frame of reference thatwas moving horizontally at a speed of one box per unit time relativeto the first frame.

12 The figure shows the motion of a point on the rim of a rollingwheel. (The shape is called a cycloid.) Suppose bug A is riding onthe rim of the wheel on a bicycle that is rolling, while bug B is onthe spinning wheel of a bike that is sitting upside down on the floor.Bug A is moving along a cycloid, while bug B is moving in a circle.Both wheels are doing the same number of revolutions per minute.Which bug has a harder time holding on, or do they find it equally

Problems 71

difficult? . Solution, p. 1039

Key to symbols:easy typical challenging difficult very difficult√

An answer check is available at www.lightandmatter.com.


a / James Joule, 1818-1889.The son of a wealthy brewer,Joule was tutored as a youngman by the famous scientist JohnDalton. Fascinated by electricity,he and his brother experimentedby giving electric shocks to eachother and to the family’s servants.Joule ran the brewery as an adult,and science was merely a serioushobby. His work on energy canbe traced to his attempt to buildan electric motor that wouldreplace steam engines. His ideaswere not accepted at first, partlybecause they contradicted thewidespread belief that heat wasa fluid, and partly because theydepended on extremely precisemeasurements, which had notpreviously been common inphysics.

Chapter 2

Conservation of Energy

Do you pronounce it Joule’s to rhyme with schools,Joule’s to rhyme with Bowls,or Joule’s to rhyme with Scowls?Whatever you call it, by Joule’s,or Joule’s,or Joule’s, it’s good!

Advertising slogan of the Joule brewery. The name, and thecorresponding unit of energy, are now usually pronounced soas to rhyme with “school.”

2.1 Energy2.1.1 The energy concept

You’d probably like to be able to drive your car and light yourapartment without having to pay money for gas and electricity, andif you do a little websurfing, you can easily find people who saythey have the solution to your problem. This kind of scam has beenaround for centuries. It used to be known as a perpetual motionmachine, but nowadays the con artists’ preferred phrase is “freeenergy.”1 A typical “free-energy” machine would be a sealed boxthat heats your house without needing to be plugged into a wallsocket or a gas pipe. Heat comes out, but nothing goes in, and thiscan go on indefinitely. But an interesting thing happens if you tryto check on the advertised performance of the machine. Typically,you’ll find out that either the device is still in development, or it’sback-ordered because so many people have already taken advantageof this Fantastic Opportunity! In a few cases, the magic box exists,but the inventor is only willing to demonstrate very small levels ofheat output for short periods of time, in which case there’s probablya tiny hearing-aid battery hidden in there somewhere, or some othertrick.

Since nobody has ever succeeded in building a device that createsheat out of nothing, we might also wonder whether any device exists

1An entertaining account of this form of quackery is given in Voodoo Sci-ence: The Road from Foolishness to Fraud, Robert Park, Oxford Univer-sity Press, 2000. Until reading this book, I hadn’t realized the degree to whichpseudoscience had penetrated otherwise respectable scientific organizations likeNASA.

73

b / Heat energy can be con-verted to light energy. Very hotobjects glow visibly, and evenobjects that aren’t so hot giveoff infrared light, a color of lightthat lies beyond the red end ofthe visible rainbow. This photowas made with a special camerathat records infrared light. Theman’s warm skin emits quite abit of infrared light energy, whilehis hair, at a lower temperature,emits less.

that can do the opposite, turning heat into nothing. You might thinkthat a refrigerator was such a device, but actually your refrigeratordoesn’t destroy the heat in the food. What it really does is to extractsome of the heat and bring it out into the room. That’s why it hasbig radiator coils on the back, which get hot when it’s in operation.

If it’s not possible to destroy or create heat outright, then youmight start to suspect that heat was a conserved quantity. Thiswould be a successful rule for explaining certain processes, such asthe transfer of heat between a cold Martini and a room-temperatureolive: if the olive loses a little heat, then the drink must gain thesame amount. It would fail in general, however. Sunlight can heatyour skin, for example, and a hot lightbulb filament can cool off byemitting light. Based on these observations, we could revise our pro-posed conservation law, and say that there is something called heat-pluslight, which is conserved. Even this, however, needs to be gen-eralized in order to explain why you can get a painful burn playingbaseball when you slide into a base. Now we could call it heatplus-lightplusmotion. The word is getting pretty long, and we haven’teven finished the list.

Rather than making the word longer and longer, physicists havehijacked the word “energy” from ordinary usage, and give it a new,specific technical meaning. Just as the Parisian platinum-iridiumkilogram defines a specific unit of mass, we need to pick somethingthat defines a definite unit of energy. The metric unit of energy isthe joule (J), and we’ll define it as the amount of energy requiredto heat 0.24 grams of water from 20 to 21 degrees Celsius. (Don’tmemorize the numbers.)2

Temperature of a mixture example 1. If 1.0 kg of water at 20◦C is mixed with 4.0 kg of water at 30◦C,what is the temperature of the mixture?

. Let’s assume as an approximation that each degree of tem-perature change corresponds to the same amount of energy. Inother words, we assume ∆E=mc∆T , regardless of whether, asin the definition of the joule, we have ∆T = 21◦C-20◦C or, as inthe present example, some other combination of initial and finaltemperatures. To be consistent with the definition of the joule, wemust have c = (1 J)/(0.24 g)/(1◦C) = 4.2×103 J/kg·◦C, which isreferred to as the specific heat of water.

2Although the definition refers to the Celsius scale of temperature, it’s notnecessary to give an operational definition of the temperature concept in general(which turns out to be quite a tricky thing to do completely rigorously); weonly need to establish two specific temperatures that can be reproduced on ther-mometers that have been calibrated in a standard way. Heat and temperatureare discussed in more detail in section 2.4, and in chapter 5. Conceptually, heatis a measure of energy, whereas temperature relates to how concentrated thatenergy is.

74 Chapter 2 Conservation of Energy

Conservation of energy tells us ∆E = 0, so

m1c∆T1 + m2c∆T2 = 0

or

∆T1

∆T2= −m2

m1

= −4.0 .

If T1 has to change four times as much as T2, and the two fi-nal temperatures are equal, then the final temperature must be28◦C.

Note how only differences in temperature and energy appearedin the preceding example. In other words, we don’t have to makeany assumptions about whether there is a temperature at which allan object’s heat energy is removed. Historically, the energy andtemperature units were invented before it was shown that there issuch a temperature, called absolute zero. There is a scale of tem-perature, the Kelvin scale, in which the unit of temperature is thesame as the Celsius degree, but the zero point is defined as absolutezero. But as long as we only deal with temperature differences, itdoesn’t matter whether we use Kelvin or Celsius. Likewise, as longas we deal with differences in heat energy, we don’t normally haveto worry about the total amount of heat energy the object has. Instandard physics terminology, “heat” is used only to refer to differ-ences, while the total amount is called the object’s “thermal energy.”This distinction is often ignored by scientists in casual speech, andin this book I’ll usually use “heat” for either quantity.

We’re defining energy by adding up things from a list, which welengthen as needed: heat, light, motion, etc. One objection to thisapproach is aesthetic: physicists tend to regard complication as asynonym for ugliness. If we have to keep on adding more and moreforms of energy to our laundry list, then it’s starting to sound likeenergy is distressingly complicated. Luckily it turns out that energyis simpler than it seems. Many forms of energy that are apparentlyunrelated turn out to be manifestations of a small number of formsat the atomic level, and this is the topic of section 2.4.


A The ancient Greek philosopher Aristotle said that objects “naturally”tended to slow down, unless there was something pushing on them tokeep them moving. What important insight was he missing?

2.1.2 Logical issues

Another possible objection is that the open-ended approach todefining energy might seem like a kind of cheat, since we keep oninventing new forms whenever we need them. If a certain experi-ment seems to violate conservation of energy, can’t we just invent

Section 2.1 Energy 75

c / As in figure b, an infraredcamera distinguishes hot andcold areas. As the bike skids toa stop with its brakes locked, thekinetic energy of the bike andrider is converted into heat inboth the floor (top) and the tire(bottom).

a new form of invisible “mystery energy” that patches things up?This would be like balancing your checkbook by putting in a faketransaction that makes your calculation of the balance agree withyour bank’s. If we could fudge this way, then conservation of energywould be untestable — impossible to prove or disprove.

Actually all scientific theories are unprovable. A theory cannever be proved, because the experiments can only cover a finitenumber out of the infinitely many situations in which the theory issupposed to apply. Even a million experiments won’t suffice to proveit in the same sense of the word “proof” that is used in mathematics.However, even one experiment that contradicts a theory is sufficientto show that the theory is wrong. A theory that is immune todisproof is a bad theory, because there is no way to test it. Forinstance, if I say that 23 is the maximum number of angels thatcan dance on the head of a pin, I haven’t made a properly falsifiablescientific theory, since there’s no method by which anyone could evenattempt to prove me wrong based on observations or experiments.

Conservation of energy is testable because new forms of energyare expected to show regular mathematical behavior, and are sup-posed to be related in a measurable way to observable phenomena.As an example, let’s see how to extend the energy concept to includemotion.

2.1.3 Kinetic energy

Energy of motion is called kinetic energy . (The root of the wordis the same as the word “cinema” – in French, kinetic energy is“energie cinetique.”) How does an object’s kinetic energy dependon its mass and velocity? Joule attempted a conceptually simpleexperiment on his honeymoon in the French-Swiss Alps near Mt.Chamonix, in which he measured the difference in temperature be-tween the top and bottom of a waterfall. The water at the top ofthe falls has some gravitational energy, which isn’t our subject rightnow, but as it drops, that gravitational energy is converted into ki-netic energy, and then into heat energy due to internal friction inthe churning pool at the bottom:

gravitational energy→ kinetic energy→ heat energy

In the logical framework of this book’s presentation of energy, thesignificance of the experiment is that it provides a way to find outhow an object’s kinetic energy depends on its mass and velocity. Theincrease in heat energy should equal the kinetic energy of the waterjust before impact, so in principle we could measure the water’smass, velocity, and kinetic energy, and see how they relate to oneanother.3

3From Joule’s point of view, the point of the experiment was different. Atthat time, most physicists believed that heat was a quantity that was conserved


d / A simplified drawing ofJoule’s paddlewheel experiment.

e / The heating of the tireand floor in figure c is somethingthat the average person mighthave predicted in advance, butthere are other situations whereit’s not so obvious. When aball slams into a wall, it doesn’trebound with the same amount ofkinetic energy. Was some energydestroyed? No. The ball andthe wall heat up. These infraredphotos show a squash ball atroom temperature (top), and afterit has been played with for severalminutes (bottom), causing it toheat up detectably.

Although the story is picturesque and memorable, most booksthat mention the experiment fail to note that it was a failure! Theproblem was that heat wasn’t the only form of energy being released.In reality, the situation was more like this:

gravitational energy→kinetic energy

→heat energy

+ sound energy

+ energy of partial evaporation.

The successful version of the experiment, shown in figures d andf, used a paddlewheel spun by a dropping weight. As with thewaterfall experiment, this one involves several types of energy, butthe difference is that in this case, they can all be determined andtaken into account. (Joule even took the precaution of putting ascreen between himself and the can of water, so that the infraredlight emitted by his warm body wouldn’t warm it up at all!) Theresult4 is

K =1

2mv2 [kinetic energy].

Whenever you encounter an equation like this for the first time,you should get in the habit of interpreting it. First off, we can tellthat by making the mass or velocity greater, we’d get more kineticenergy. That makes sense. Notice, however, that we have mass tothe first power, but velocity to the second. Having the whole thingproportional to mass to the first power is necessary on theoreticalgrounds, since energy is supposed to be additive. The dependenceon v2 couldn’t have been predicted, but it is sensible. For instance,suppose we reverse the direction of motion. This would reverse thesign of v, because in one dimension we use positive and negative signsto indicate the direction of motion. But since v2 is what appears inthe equation, the resulting kinetic energy is unchanged.

separately from the rest of the things to which we now refer as energy, i.e.,mechanical energy. Separate units of measurement had been constructed forheat and mechanical of energy, but Joule was trying to show that one couldconvert back and forth between them, and that it was actually their sum thatwas conserved, if they were both expressed in consistent units. His main resultwas the conversion factor that would allow the two sets of units to be reconciled.By showing that the conversion factor came out the same in different typesof experiments, he was supporting his assertion that heat was not separatelyconserved. From Joule’s perspective or from ours, the result is to connect themysterious, invisible phenomenon of heat with forms of energy that are visibleproperties of objects, i.e., mechanical energy.

4If you’ve had a previous course in physics, you may have seen this presentednot as an empirical result but as a theoretical one, derived from Newton’s laws,and in that case you might feel you’re being cheated here. However, I’m go-ing to reverse that reasoning and derive Newton’s laws from the conservationlaws in chapter 3. From the modern perspective, conservation laws are morefundamental, because they apply in cases where Newton’s laws don’t.


f / A realistic drawing of Joule’s apparatus, based on the illustrationin his original paper. The paddlewheel is sealed inside the can in themiddle. Joule wound up the two 13-kg lead weights and dropped them1.6 meters, repeating this 20 times to produce a temperature changeof only about half a degree Fahrenheit in the water inside the sealedcan. He claimed in his paper to be able to measure temperatures to anaccuracy of 1/200 of a degree.

What about the factor of 1/2 in front? It comes out to beexactly 1/2 by the design of the metric system. If we’d been usingthe old-fashioned British engineering system of units (which is nolonger used in the U.K.), the equation would have been K = (7.44×10−2 Btu · s2/slug · ft2)mv2. The version of the metric system calledthe SI,5 in which everything is based on units of kilograms, meters,and seconds, not only has the numerical constant equal to 1/2, butmakes it unitless as well. In other words, we can think of the jouleas simply an abbreviation, 1 J=1 kg ·m2/s2. More familiar examplesof this type of abbreviation are 1 minute=60 s, and the metric unitof land area, 1 hectare=10000 m2.

Ergs and joules example 2. There used to be two commonly used systems of metric units,referred to as mks and cgs. The mks system, now called the SI,is based on the meter, the kilogram, and the second. The cgssystem, which is now obsolete, was based on the centimeter, thegram, and the second. In the cgs system, the unit of energy isnot the joule but the erg, 1 erg=1 g · cm2/s2. How many ergs arein one joule?

. The simplest approach is to treat the units as if they were alge-

5Systeme International


bra symbols.

1 J = 1kg ·m2

s2

= 1kg ·m2

s2 × 1000 g1 kg

×(

100 cm1 m

)2

= 107 g · cm2

s2

= 107 erg

Cabin air in a jet airplane example 3. A jet airplane typically cruises at a velocity of 270 m/s. Outsideair is continuously pumped into the cabin, but must be cooled offfirst, both because (1) it heats up due to friction as it enters the en-gines, and (2) it is heated as a side-effect of being compressed tocabin pressure. Calculate the increase in temperature due to thefirst effect. The specific heat of dry air is about 1.0×103 J/kg·◦C.

. This is easiest to understand in the frame of reference of theplane, in which the air rushing into the engine is stopped, and itskinetic energy converted into heat.6 Conservation of energy tellsus

0 = ∆E= ∆K + ∆Eheat .

In the plane’s frame of reference, the air’s initial velocity is vi=270m/s, and its final velocity is zero, so the change in its kinetic en-ergy is negative,

∆K = Kf − Ki

= 0− (1/2)mvi2

= −(1/2)mvi2.

Assuming that the specific heat of air is roughly independent oftemperature (which is why the number was stated with the word“about”), we can substitute into 0 = ∆K + ∆Eheat , giving

0 = −12

mvi2 + mc∆T

12

vi2 = c∆T .

Note how the mass cancels out. This is a big advantage of solvingproblems algebraically first, and waiting until the end to plug in6It’s not at all obvious that the solution would work out in the earth’s frame of

reference, although Galilean relativity states that it doesn’t matter which framewe use. Chapter 3 discusses the relationship between conservation of energy andGalilean relativity.


numbers. With a purely numerical approach, we wouldn’t evenhave known what value of m to pick, or if we’d guessed a valuelike 1 kg, we wouldn’t have known whether our answer dependedon that guess.

Solving for ∆T , and writing v instead of vi for simplicity, we find

∆T =v2

2c≈ 40◦C.

The passengers would be boiled alive if not for the refrigeration.The first stage of cooling happens via heat exchangers in theengine struts, but a second stage, using a refrigerator under thefloor of the cabin, is also necessary. Running this refrigeratoruses up energy, cutting into the fuel efficiency of the airplane,which is why typically only 50% of the cabin’s air is replaced ineach pumping cycle of 2-3 minutes. Although the airlines preferto emphasize that this is a much faster recirculation rate than inthe ventilation systems of most buildings, people are packed moretightly in an airplane.

2.1.4 Power

Power, P , is defined as the rate of change of energy, dE/dt.Power thus has units of joules per second, which are usually abbre-viated as watts, 1 W=1 J/s. Since energy is conserved, we wouldhave dE/dt = 0 if E was the total energy of a closed system, andthat’s not very interesting. What’s usually more interesting to dis-cuss is either the power flowing in or out of an open system, or therate at which energy is being transformed from one form into an-other. The following is an example of energy flowing into an opensystem.

Heating by a lightbulb example 4. The electric company bills you for energy in units of kilowatt-hours (kilowatts multiplied by hours) rather than in SI units ofjoules. How many joules is a kilowatt-hour?

. 1 kilowatt-hour = (1 kW)(1 hour) = (1000 J/s)(3600 s) = 3.6 MJ.

Now here’s an example of energy being transformed from oneform into another.

Human wattage example 5. Food contains chemical energy (discussed in more detail in sec-tion 2.4), and for historical reasons, food energy is normally givenin non-SI units of Calories. One Calorie with a capital “C” equals1000 calories, and 1 calorie is defined as 4.18 J. A typical personconsumes 2000 Calories of food in a day, and converts nearly allof that directly to body heat. Compare the person’s heat produc-tion to the rate of energy consumption of a 100-watt lightbulb.

. Strictly speaking, we can’t really compute the derivative dE/dt ,


g / A skateboarder rises tothe edge of an empty pool andthen falls back down.

h / The sum of kinetic plusgravitational energy is constant.

since we don’t know how the person’s metabolism ebbs and flowsover the course of a day. What we can really compute is ∆E/∆t ,which is the power averaged over a one-day period.

Converting to joules, we find ∆E = 8 × 106 J for the amount ofenergy transformed into heat within our bodies in one day. Con-verting the time interval likewise into SI units, ∆t = 9 × 104 s.Dividing, we find that our power is 90 J/s = 90 W, about the sameas a lightbulb.

2.1.5 Gravitational energy

Gravitational energy, to which I’ve already alluded, is differentfrom heat and kinetic energy in an important way. Heat and kineticenergy are properties of a single object, whereas gravitational energydescribes an interaction between two objects. When the skater infigures g and h is at the top, his distance from the bulk of the planetearth is greater. Since we observe his kinetic energy decreasing onthe way up, there must be some other form of energy that is increas-ing. We invent a new form of energy, called gravitational energy, andwritten U or Ug, which depends on the distance between his bodyand the planet. Where is this energy? It’s not in the skater’s body,and it’s not inside the earth, either, since it takes two to tango. Ifeither object didn’t exist, there wouldn’t be any interaction or anyway to measure a distance, so it wouldn’t make sense to talk abouta distance-dependent energy. Just as marriage is a relationship be-tween two people, gravitational energy is a relationship between twoobjects.

There is no precise way to define the distance between the skaterand the earth, since both are objects that have finite size. As dis-cussed in more detail in section 2.3, gravity is one of the fundamen-tal forces of nature, a universal attraction between any two particlesthat have mass. Each atom in the skater’s body is at a definite dis-tance from each atom in the earth, but each of these distances isdifferent. An atom in his foot is only a few centimeters from someof the atoms in the plaster side of the pool, but most of the earth’satoms are thousands of kilometers away from him. In theory, wemight have to add up the contribution to the gravitational energyfor every interaction between an atom in the skater’s body and anatom in the earth.

For our present purposes, however, there is a far simpler andmore practical way to solve problems. In any region of the earth’ssurface, there is a direction called “down,” which we can establishby dropping a rock or hanging a plumb bob. In figure h, the skater ismoving up and down in one dimension, and if we did measurementsof his kinetic energy, like the made-up data in the figure, we couldinfer his gravitational energy. As long as we stay within a relativelysmall range of heights, we find that an object’s gravitational energyincreases at a steady rate with height. In other words, the strength


of gravity doesn’t change much if you only move up or down a fewmeters. We also find that the gravitational energy is proportionalto the mass of the object we’re testing. Writing y for the height,and g for the overall constant of proportionality, we have

Ug = mgy. [gravitational energy; y=height; only ac-curate within a small range of heights]

The number g, with units of joules per kilogram per meter, is calledthe gravitational field . It tells us the strength of gravity in a certainregion of space. Near the surface of our planet, it has a value ofabout 9.8 J/kg ·m, which is conveniently close to 10 J/kg ·m forrough calculations.

Velocity at the bottom of a drop example 6. If the skater in figure g drops 3 meters from rest, what is hisvelocity at the bottom of the pool?

. Starting from conservation of energy, we have

0 = ∆E= ∆K + ∆U= Kf − Ki + Uf − Ui

=12

mvf2 + mgyf −mgyi (because Ki=0)

=12

mvf2 + mg∆y , (∆y <0)

so

v =√−2g∆y

=√−(2)(10 J/kg·m)(− 3 m)

= 8 m/s (rounded to one sig. fig.)

There are a couple of important things to note about this ex-ample. First, we were able to massage the equation so that it onlyinvolved ∆y, rather than y itself. In other words, we don’t need toworry about where y = 0 is; any coordinate system will work, aslong as the positive y axis points up, not down. This is no accident.Gravitational energy can always be changed by adding a constantonto it, with no effect on the final result, as long as you’re consistentwithin a given problem.

The other interesting thing is that the mass canceled out: evenif the skater gained weight or strapped lead weights to himself, hisvelocity at the bottom would still be 8 m/s. This isn’t an accidenteither. This is the same conclusion we reached in section 1.2, basedon the equivalence of gravitational and inertial mass. The kineticenergy depends on the inertial mass, while gravitational energy is


related to gravitational mass, but since these two masses are equal,we were able to use a single symbol, m, for them, and cancel themout.

We can see from the equation v =√−2g∆y that a falling object’s

velocity isn’t constant. It increases as the object drops farther andfarther. What about its acceleration? If we assume that air frictionis negligible, the arguments in section 1.2 show that the accelera-tion can’t depend on the object’s mass, so there isn’t much else theacceleration can depend on besides g. In fact, the acceleration of afalling object equals −g (in a coordinate system where the positivey axis points up), as we can easily show using the chain rule:(

dv

dt

)=

(dv

dK

)(dK

dU

)(dU

dy

)(dy

dt

)=

(1

mv

)(−1)(mg)(v)

= −g,

where I’ve calculated dv/dK as 1/(dK/dv), and dK/dU = −1 canbe found by differentiatingK+U = (constant) to give dK+dU = 0.7

We can also check that the units of g, J/kg·m, are equivalent tothe units of acceleration,

J

kg·m=

kg·m2/s2

kg·m=

m

s2,

and therefore the strength of the gravitational field near the earth’ssurface can just as well be stated as 10 m/s2.

Speed after a given time example 7. An object falls from rest. How fast is it moving after two sec-onds? Assume that the amount of energy converted to heat byair friction is negligible.

. Under the stated assumption, we have a = −g, which can beintegrated to give v = −gt + constant. If we let t = 0 be thebeginning of the fall, then the constant of integration is zero, so att = 2 s we have v = −gt = −(10 m/s2)× (2 s) = 20 m/s.

The Vomit Comet example 8. The U.S. Air Force has an airplane, affectionately known as theVomit Comet, in which astronaut trainees can experience sim-ulated weightlessness. Oversimplifying a little, imagine that theplane climbs up high, and then drops straight down like a rock.(It actually flies along a parabola.) Since the people are fallingwith the same acceleration as the plane, the sensation is just likewhat you’d experience if you went out of the earth’s gravitational7There is a mathematical loophole in this argument that would allow the

object to hover for a while with zero velocity and zero acceleration. This pointis discussed on page 1025.


i / Two balls start from rest,and roll from A to B by differentpaths.

field. If the plane can start from 10 km up, what is the maximumamount of time for which the dive can last?

. Based on data about acceleration and distance, we want tofind time. Acceleration is the second derivative of distance, so ifwe integrate the acceleration twice with respect to time, we canfind how position relates to time. For convenience, let’s pick acoordinate system in which the positive y axis is down, so a=ginstead of −g.

a = gv = gt + constant (integrating)

= gt (starts from rest)

y =12

gt2 + constant (integrating again)

Choosing our coordinate system to have y = 0 at t = 0, we canmake the second constant of integration equal zero as well, so

t =

√2yg

=

√2 · 10000 m

10 m/s2

=√

2000 s2

= 40 s (to one sig. fig.)

Note that if we hadn’t converted the altitude to units of meters,we would have gotten the wrong answer, but we would have beenalerted to the problem because the units inside the square rootwouldn’t have come out to be s2. In general, it’s a good idea toconvert all your data into SI (meter-kilogram-second) units beforeyou do anything with them.

High road, low road example 9. In figure i, what can you say based on conservation of energyabout the speeds of the balls when the reach point B? What doesconservation of energy tell you about which ball will get therefirst? Assume friction doesn’t convert any mechanical energy toheat or sound energy.

. Since friction is assumed to be negligible, there are only twoforms of energy involved: kinetic and gravitational. Since bothballs start from rest, and both lose the same amount of gravi-tational energy, they must have the same kinetic energy at theend, and therefore they’re rolling at the same speed when theyreach B. (A subtle point is that the balls have kinetic energy bothbecause they’re moving through space and because they’re spin-ning as they roll. These two types of energy must be in fixed


j / How much energy is re-quired to raise the submergedbox through a height ∆y?

k / A seesaw.

l / The biceps muscle is areversed lever.

proportion to one another, so this has no effect on the conclu-sion.)

Conservation of energy does not, however, tell us anything obvi-ous about which ball gets there first. This is a general problemwith applying conservation laws: conservation laws don’t refer di-rectly to time, since they are statements that something stays thesame at all moments in time. We expect on intuitive grounds thatthe ball that goes by the lower ramp gets to B first, since it buildsup speed early on.

Buoyancy example 10. A cubical box with mass m and volume V = b3 is submerged

in a fluid of density ρ. How much energy is required to raise itthrough a height ∆y?

. As the box moves up, it invades a volume V ′ = b2∆y previouslyoccupied by some of the fluid, and fluid flows into an equal volumethat it has vacated on the bottom. Lowering this amount of fluidby a height b reduces the fluid’s gravitational energy by ρV ′gb =ρgb3∆y , so the net change in energy is

∆E = mg∆y − ρgb3∆y= (m − ρV )g∆y .

In other words, it’s as if the mass of the box had been reduced byan amount equal to the fluid that otherwise would have occupiedthat volume. This is known as Archimedes’ principle, and it istrue even if the box is not a cube, although we’ll defer the moregeneral proof until page 207 in chapter 3. If the box is less densethan the fluid, then it will float.

A simple machine example 11. If the father and son on the seesaw in figure k start from rest,what will happen?

. Note that although the father is twice as massive, he is at halfthe distance from the fulcrum. If the seesaw was going to startrotating, it would have to be losing gravitational energy in order togain some kinetic energy. However, there is no way for it to gainor lose gravitational energy by rotating in either direction. Thechange in gravitational energy would be

∆U = ∆U1 + ∆U2

= g(m1∆y1 + m2∆y2),

but ∆y1 and ∆y2 have opposite signs and are in the proportion oftwo to one, since the son moves along a circular arc that coversthe same angle as the father’s but has half the radius. Therefore∆U = 0, and there is no way for the seesaw to trade gravitationalenergy for kinetic.


m / Discussion question C.

n / A hydraulic jack.

o / The surfaces are friction-less. The black blocks are inequilibrium.

The seesaw example demonstrates the principle of the lever,which is one of the basic mechanical building blocks known as sim-ple machines. As discussed in more detail in chapters 3 and 4, theprinciple applies even when the interactions involved aren’t gravita-tional.

Note that although a lever makes it easier to lift a heavy weight,it also decreases the distance traveled by the load. By reversing thelever, we can make the load travel a greater distance, at the expenseof increasing the amount of force required. The human muscular-skeletal system uses reversed levers of this kind, which allows us tomove more rapidly, and also makes our bodies more compact, at theexpense of brute strength. A piano uses reversed levers so that asmall amount of motion of the key produces a longer swing of thehammer. Another interesting example is the hydraulic jack shownin figure n. The analysis in terms of gravitational energy is exactlythe same as for the seesaw, except that the relationship between ∆y1

and ∆y2 is now determined not by geometry but by conservationof mass: since water is highly incompressible, conservation of massis approximately the same as a requirement of constant volume,which can only be satisfied if the distance traveled by each piston isin inverse proportion to its cross-sectional area.


A Hydroelectric power (water flowing over a dam to spin turbines)appears to be completely free. Does this violate conservation of energy?If not, then what is the ultimate source of the electrical energy producedby a hydroelectric plant?

B You throw a steel ball up in the air. How can you prove based onconservation of energy that it has the same speed when it falls back intoyour hand? What if you threw a feather up? Is energy not conserved inthis case?

C Figure m shows a pendulum that is released at A and caught by apeg as it passes through the vertical, B. To what height will the bob riseon the right?

D What is wrong with the following definitions of g?(a) “g is gravity.”(b) “g is the speed of a falling object.”(c) “g is how hard gravity pulls on things.”

2.1.6 Equilibrium and stability

The seesaw in figure k is in equilibrium, meaning that if it startsout being at rest, it will stay put. This is known as a neutral equi-librium, since the seesaw has no preferred position to which it willreturn if we disturb it. If we move it to a different position andrelease it, it will stay at rest there as well. If we put it in motion, itwill simply continue in motion until one person’s feet hit the ground.

Most objects around you are in stable equilibria, like the black


block in figure o/3. Even if the block is moved or set in motion, itwill oscillate about the equilibrium position. The pictures are likegraphs of y versus x, but since the gravitational energy U = mgyis proportional to y, we can just as well think of them as graphsof U versus x. The block’s stable equilibrium position is wherethe function U(x) has a local minimum. The book you’re readingright now is in equilibrium, but gravitational energy isn’t the onlyform of energy involved. To move it upward, we’d have to supplygravitational energy, but downward motion would require a differentkind of energy, in order to compress the table more. (As we’ll seein section 2.4, this is electrical energy due to interactions betweenatoms within the table.)

A differentiable function’s local extrema occur where its deriva-tive is zero. A position where dU/ dx is zero can be a stable (3),neutral (2), or unstable equilibrium, (4). An unstable equilibriumis like a pencil balanced on its tip. Although it could theoreticallyremain balanced there forever, in reality it will topple due to anytiny perturbation, such as an air current or a vibration from a pass-ing truck. This is a technical, mathematical definition of instability,which is more restrictive than the way the word is used in ordinaryspeech. Most people would describe a domino standing upright asbeing unstable, but in technical usage it would be considered stable,because a certain finite amount of energy is required to tip it over,and perturbations smaller than that would only cause it to oscillatearound its equilibrium position.

The domino is also an interesting example because it has twolocal minima, one in which it is upright, and another in which it islying flat. A local minimum that is not the global minimum, as infigure o/5, is referred to as a metastable equilibrium.

p / Example 12.

A neutral equilibrium example 12Figure p shows a special-purpose one-block funicular railroadnear Hill and Fourth Streets in Los Angeles, California, used forgetting passengers up and down a very steep hill. It has two carsattached to a single loop of cable, arranged so that while one cargoes up, the other comes down. They pass each other in themiddle. Since one car’s gravitational energy is increasing while


the other’s is decreasing, the system is in neutral equilibrium. Ifthere were no frictional heating, exactly zero energy would be re-quired in order to operate the system. A similar counterweightingprinciple is used in aerial tramways in mountain resorts, and inelevators (with a solid weight, rather than a second car, as coun-terweight).


q / Water in a U-shaped tube.

Water in a U-shaped tube example 13. The U-shaped tube in figure q has cross-sectional area A, andthe density of the water inside is ρ. Find the gravitational energyas a function of the quantity y shown in the figure, and show thatthere is an equilibrium at y=0.

. The question is a little ambiguous, since gravitational energyis only well defined up to an additive constant. To fix this con-stant, let’s define U to be zero when y=0. The difference betweenU(y ) and U(0) is the energy that would be required to lift a wa-ter column of height y out of the right side, and place it abovethe dashed line, on the left side, raising it through a height y .This water column has height y and cross-sectional area A, soits volume is Ay , its mass is ρAy , and the energy required ismgy=(ρAy )gy=ρgAy2. We then have U(y ) = U(0) + ρgAy2 =ρgAy2.

To find equilibria, we look for places where the derivative dU/dy =2ρgAy equals 0. As we’d expect intuitively, the only equilibriumoccurs at y=0. The second derivative test shows that this is alocal minimum (not a maximum or a point of inflection), so this isa stable equilibrium.

2.1.7 Predicting the direction of motion

Kinetic energy doesn’t depend on the direction of motion. Some-times this is helpful, as in the high road-low road example (p. 84,example 9), where we were able to predict that the balls would havethe same final speeds, even though they followed different paths andwere moving in different directions at the end. In general, however,the two conservation laws we’ve encountered so far aren’t enough topredict an object’s path through space, for which we need conserva-tion of momentum (chapter 3), and the mathematical technique ofvectors. Before we develop those ideas in their full generality, how-ever, it will be helpful to do a couple of simple examples, includingone that we’ll get a lot of mileage out of in section 2.3.

Suppose we observe an air hockey puck gliding frictionlessly tothe right at a velocity v, and we want to predict its future motion.Since there is no friction, no kinetic energy is converted to heat.The only form of energy involved is kinetic energy, so conservationof energy, ∆E = 0, becomes simply ∆K = 0. There’s no particularreason for the puck to do anything but continue moving to the rightat constant speed, but it would be equally consistent with conserva-tion of energy if it spontaneously decided to reverse its direction ofmotion, changing its velocity to −v. Either way, we’d have ∆K = 0.There is, however, a way to tell which motion is physical and whichis unphysical. Suppose we consider the whole thing again in theframe of reference that is initially moving right along with the puck.In this frame, the puck starts out with K = 0. What we origi-nally described as a reversal of its velocity from v to −v is, in this


r / A car drives over a cliff.

new frame of reference, a change from zero velocity to −2v, whichwould violate conservation of energy. In other words, the physicallypossible motion conserves energy in all frames of reference, but theunphysical motion only conserves energy in one special frame ofreference.

For our second example, we consider a car driving off the edgeof a cliff (r). For simplicity, we assume that air friction is negligible,so only kinetic and gravitational energy are involved. Does the carfollow trajectory 1, familiar from Road Runner cartoons, trajectory2, a parabola, or 3, a diagonal line? All three could be consistentwith conservation of energy, in the ground’s frame of reference. Forinstance, the car would have constant gravitational energy alongthe initial horizontal segment of trajectory 1, so during that timeit would have to maintain constant kinetic energy as well. Only aparabola, however, is consistent with conservation of energy com-bined with Galilean relativity. Consider the frame of reference thatis moving horizontally at the same speed as that with which the carwent over the edge. In this frame of reference, the cliff slides outfrom under the initially motionless car. The car can’t just hoverfor a while, so trajectory 1 is out. Repeating the same math as inexample 8 on p. 83, we have

x∗ = 0 , y∗ =1

2gt2

in this frame of reference, where the stars indicate coordinates mea-sured in the moving frame of reference. These coordinates are re-lated to the ground-fixed coordinates (x, y) by the equations

x = x∗ + vt and y = y∗,

where v is the velocity of one frame with respect to the other. Wetherefore have

x = vt , y =1

2gt2,

in our original frame of reference. Eliminating t, we can see thatthis has the form of a parabola:

y =g

2v2x2.

self-check AWhat would the car’s motion be like in the * frame of reference if itfollowed trajectory 3? . Answer, p. 1058


2.2 Numerical techniquesEngineering majors are a majority of the students in the kind ofphysics course for which this book is designed, so most likely youfall into that category. Although you surely recognize that physicsis an important part of your training, if you’ve had any exposureto how engineers really work, you’re probably skeptical about theflavor of problem-solving taught in most science courses. You real-ize that not very many practical engineering calculations fall intothe narrow range of problems for which an exact solution can becalculated with a piece of paper and a sharp pencil. Real-life prob-lems are usually complicated, and typically they need to be solvedby number-crunching on a computer, although we can often gaininsight by working simple approximations that have algebraic solu-tions. Not only is numerical problem-solving more useful in real life,it’s also educational; as a beginning physics student, I really onlyfelt like I understood projectile motion after I had worked it bothways, using algebra and then a computer program. (This was backin the days when 64 kilobytes of memory was considered a lot.)

In this section, we’ll start by seeing how to apply numericaltechniques to some simple problems for which we know the answer in“closed form,” i.e., a single algebraic expression without any calculusor infinite sums. After that, we’ll solve a problem that would havemade you world-famous if you could have done it in the seventeenthcentury using paper and a quill pen! Before you continue, you shouldread Appendix 1 on page 1023 that introduces you to the Pythonprogramming language.

First let’s solve the trivial problem of finding how much time ittakes an object moving at speed v to travel a straight-line distancedist. This closed-form answer is, of course, dist/v, but the pointis to introduce the techniques we can use to solve other problems ofthis type. The basic idea is to divide the distance up into n equalparts, and add up the times required to traverse all the parts. Thefollowing Python function does the job. Note that you shouldn’ttype in the line numbers on the left, and you don’t need to typein the comments, either. I’ve omitted the prompts >>> and ... inorder to save space.

1 import math

2 def time1(dist,v,n):

3 x=0 # Initialize the position.

4 dx = dist/n # Divide dist into n equal parts.

5 t=0 # Initialize the time.

6 for i in range(n):

7 x = x+dx # Change x.

8 dt=dx/v # time=distance/speed

9 t=t+dt # Keep track of elapsed time.

10 return t

Section 2.2 Numerical techniques 91

11

How long does it take to move 1 meter at a constant speed of 1 m/s?If we do this,

>>> print(time1(1.0,1.0,10)) # dist, v, n

0.99999999999999989

Python produces the expected answer by dividing the distance intoten equal 0.1-meter segments, and adding up the ten 0.1-secondtimes required to traverse each one. Since the object moves at con-stant speed, it doesn’t even matter whether we set n to 10, 1, or amillion:

>>> print(time1(1.0,1.0,1)) # dist, v, n

1.0

Now let’s do an example where the answer isn’t obvious to peoplewho don’t know calculus: how long does it take an object to fallthrough a height h, starting from rest? We know from example 8on page 83 that the exact answer, found using calculus, is

√2h/g.

Let’s see if we can reproduce that answer numerically. The maindifference between this program and the previous one is that nowthe velocity isn’t constant, so we need to update it as we go along.Conservation of energy gives mgh = (1/2)mv2+mgy for the velocityv at height y, so v = −

√2g(h− y). (We choose the negative root

because the object is moving down, and our coordinate system hasthe positive y axis pointing up.)

1 import math

2 def time2(h,n):

3 g=9.8 # gravitational field

4 y=h # Initialize the height.

5 v=0 # Initialize the velocity.

6 dy = -h/n # Divide h into n equal parts.

7 t=0 # Initialize the time.


9 y = y+dy # Change y. (Note dy<0.)

10 v = -math.sqrt(2*g*(h-y)) # from cons. of energy

11 dt=dy/v # dy and v are <0, so dt is >0

12 t=t+dt # Keep track of elapsed time.

13 return t

14

For h=1.0 m, the closed-form result is√

2 · 1.0 m/9.8 m/s2 = 0.45 s.With the drop split up into only 10 equal height intervals, the nu-merical technique provides a pretty lousy approximation:


>>> print(time2(1.0,10)) # h, n

0.35864270709233342

But by increasing n to ten thousand, we get an answer that’s asclose as we need, given the limited accuracy of the raw data:

>>> print(time2(1.0,10000)) # h, n

0.44846664060793945

A subtle point is that we changed y in line 9, and then on line10 we calculated v, which depends on y. Since y is only changingby a ten-thousandth of a meter with each step, you might thinkthis wouldn’t make much of a difference, and you’d be almost right,except for one small problem: if we swapped lines 9 and 10, thenthe very first time through the loop, we’d have v=0, which wouldproduce a division-by-zero error when we calculated dt! Actuallywhat would make the most sense would be to calculate the velocityat height y and the velocity at height y+dy (recalling that dy isnegative), average them together, and use that value of y to calculatethe best estimate of the velocity between those two points. Sincethe acceleration is constant in the present example, this modificationresults in a program that gives an exact result even for n=1:

1 import math

2 def time3(h,n):

3 g=9.8

4 y=h

5 v=0

6 dy = -h/n

7 t=0


9 y_old = y

10 y = y+dy

11 v_old = math.sqrt(2*g*(h-y_old))

12 v = math.sqrt(2*g*(h-y))

13 v_avg = -(v_old+v)/2.

14 dt=dy/v_avg

15 t=t+dt

16 return t

17

>>> print(time3(1.0,1)) # h, n

0.45175395145262565

Now we’re ready to attack a problem that challenged the bestminds of Europe back in the days when there were no computers.In 1696, the mathematician Johann Bernoulli posed the following


a / Approximations to thebrachistochrone curve usinga third-order polynomial (solidline), and a seventh-order poly-nomial (dashed). The latteronly improves the time by fourmilliseconds.

famous question. Starting from rest, an object slides frictionlesslyover a curve joining the point (a, b) to the point (0, 0). Of all thepossible shapes that such a curve could have, which one gets theobject to its destination in the least possible time, and how muchtime does it take? The optimal curve is called the brachistochrone,from the Greek “short time.” The solution to the brachistochroneproblem evaded Bernoulli himself, as well as Leibniz, who had beenone of the inventors of calculus. The English physicist Isaac Newton,however, stayed up late one night after a day’s work running theroyal mint, and, according to legend, produced an algebraic solutionat four in the morning. He then published it anonymously, butBernoulli is said to have remarked that when he read it, he knewinstantly from the style that it was Newton — he could “tell thelion from the mark of his claw.”

Rather than attempting an exact algebraic solution, as Newtondid, we’ll produce a numerical result for the shape of the curveand the minimum time, in the special case of a=1.0 m and b=1.0 m.Intuitively, we want to start with a fairly steep drop, since any speedwe can build up at the start will help us throughout the rest of themotion. On the other hand, it’s possible to go too far with this idea:if we drop straight down for the whole vertical distance, and thendo a right-angle turn to cover the horizontal distance, the resultingtime of 0.68 s is quite a bit longer than the optimal result, the reasonbeing that the path is unnecessarily long. There are infinitely manypossible curves for which we could calculate the time, but let’s lookat third-order polynomials,

y = c1x+ c2x2 + c3x

3,

where we require c3 = (b−c1a−c2a2)/a3 in order to make the curve

pass through the point (a, b). The Python program, below, is notmuch different from what we’ve done before. The function only asksfor c1 and c2, and calculates c3 internally at line 4. Since the motionis two-dimensional, we have to calculate the distance between onepoint and the next using the Pythagorean theorem, at line 16.

1 import math

2 def timeb(a,b,c1,c2,n):

3 g=9.8

4 c3 = (b-c1*a-c2*a**2)/(a**3)

5 x=a

6 y=b

7 dx = -a/n

8 t=0


10 y_old = y

11 x = x+dx

12 y = c1*x+c2*x**2+c3*x**3


13 dy = y-y_old

14 v_old = math.sqrt(2*g*(b-y_old))

15 v = math.sqrt(2*g*(b-y))

16 v_avg = (v_old+v)/2.

17 ds = math.sqrt(dx**2+dy**2) # Pythagorean thm.

18 dt=ds/v_avg

19 t=t+dt

20 return t

21

As a first guess, we could try a straight diagonal line, y = x,which corresponds to setting c1 = 1, and all the other coefficients tozero. The result is a fairly long time:

>>> a=1.

>>> b=1.

>>> n=10000

>>> c1=1.

>>> c2=0.

>>> print(timeb(a,b,c1,c2,n))

0.63887656499994161

What we really need is a curve that’s very steep on the right, andflatter on the left, so it would actually make more sense to try y = x3:

>>> c1=0.

>>> c2=0.

>>> print(timeb(a,b,c1,c2,n))

0.59458339947087069

This is a significant improvement, and turns out to be only a hun-dredth of a second off of the shortest possible time! It’s possible,although not very educational or entertaining, to find better ap-proximations to the brachistochrone curve by fiddling around withthe coefficients of the polynomial by hand. The real point of thisdiscussion was to give an example of a nontrivial problem that canbe attacked successfully with numerical techniques. I found the firstapproximation shown in figure a,

y = (0.62)x+ (−0.93)x2 + (1.31)x3

by using the program listed in appendix 2 on page 1026 to searchautomatically for the optimal curve. The seventh-order approxima-tion shown in the figure came from a straightforward extension ofthe same program.


a / An ellipse is circle that hasbeen distorted by shrinking andstretching along perpendicularaxes.

b / An ellipse can be con-structed by tying a string to twopins and drawing like this with apencil stretching the string taut.Each pin constitutes one focus ofthe ellipse.

c / If the time interval takenby the planet to move from P to Qis equal to the time interval fromR to S, then according to Kepler’sequal-area law, the two shadedareas are equal. The planetis moving faster during timeinterval RS than it was duringPQ, because gravitational energyhas been transformed into kineticenergy.

2.3 Gravitational phenomenaCruise your radio dial today and try to find any popular song thatwould have been imaginable without Louis Armstrong. By introduc-ing solo improvisation into jazz, Armstrong took apart the jigsawpuzzle of popular music and fit the pieces back together in a dif-ferent way. In the same way, Newton reassembled our view of theuniverse. Consider the titles of some recent physics books writtenfor the general reader: The God Particle, Dreams of a Fi-nal Theory. When the subatomic particle called the neutrino wasrecently proven for the first time to have mass, specialists in cos-mology began discussing seriously what effect this would have oncalculations of the evolution of the universe from the Big Bang toits present state. Without the English physicist Isaac Newton, suchattempts at universal understanding would not merely have seemedambitious, they simply would not have occurred to anyone.

This section is about Newton’s theory of gravity, which he usedto explain the motion of the planets as they orbited the sun. Newtontosses off a general treatment of motion in the first 20 pages of hisMathematical Principles of Natural Philosophy, and thenspends the next 130 discussing the motion of the planets. Clearlyhe saw this as the crucial scientific focus of his work. Why? Becausein it he showed that the same laws of nature applied to the heavensas to the earth, and that the gravitational interaction that madean apple fall was the same as the as the one that kept the earth’smotion from carrying it away from the sun.

2.3.1 Kepler’s laws

Newton wouldn’t have been able to figure out why the planetsmove the way they do if it hadn’t been for the astronomer TychoBrahe (1546-1601) and his protege Johannes Kepler (1571-1630),who together came up with the first simple and accurate descriptionof how the planets actually do move. The difficulty of their task issuggested by the figure below, which shows how the relatively simpleorbital motions of the earth and Mars combine so that as seen fromearth Mars appears to be staggering in loops like a drunken sailor.

Brahe, the last of the great naked-eye astronomers, collected ex-tensive data on the motions of the planets over a period of manyyears, taking the giant step from the previous observations’ accuracyof about 10 minutes of arc (10/60 of a degree) to an unprecedented1 minute. The quality of his work is all the more remarkable consid-ering that his observatory consisted of four giant brass protractorsmounted upright in his castle in Denmark. Four different observerswould simultaneously measure the position of a planet in order tocheck for mistakes and reduce random errors.

With Brahe’s death, it fell to his former assistant Kepler to try


d / As the earth and Mars revolvearound the sun at different rates,the combined effect of their mo-tions makes Mars appear to tracea strange, looped path across thebackground of the distant stars.

to make some sense out of the volumes of data. After 900 pages ofcalculations and many false starts and dead-end ideas, Kepler finallysynthesized the data into the following three laws:

Kepler’s elliptical orbit law: The planets orbit the sun inelliptical orbits with the sun at one focus.Kepler’s equal-area law: The line connecting a planet to thesun sweeps out equal areas in equal amounts of time.Kepler’s law of periods: The time required for a planet to orbitthe sun, called its period, T , is proportional to the long axis of theellipse raised to the 3/2 power. The constant of proportionalityis the same for all the planets.

Although the planets’ orbits are ellipses rather than circles, mostare very close to being circular. The earth’s orbit, for instance, isonly flattened by 1.7% relative to a circle. In the special case of aplanet in a circular orbit, the two foci (plural of “focus”) coincideat the center of the circle, and Kepler’s elliptical orbit law thus saysthat the circle is centered on the sun. The equal-area law impliesthat a planet in a circular orbit moves around the sun with constantspeed. For a circular orbit, the law of periods then amounts to astatement that the time for one orbit is proportional to r3/2, wherer is the radius. If all the planets were moving in their orbits at thesame speed, then the time for one orbit would simply depend onthe circumference of the circle, so it would only be proportional tor to the first power. The more drastic dependence on r3/2 meansthat the outer planets must be moving more slowly than the innerplanets.

Section 2.3 Gravitational phenomena 97

e / A cannon fires cannon-balls at different velocities, fromthe top of an imaginary mountainthat rises above the earth’s atmo-sphere. This is almost the sameas a figure Newton included in hisMathematical Principles.

Our main focus in this section will be to use the law of periodsto deduce the general equation for gravitational energy. The equal-area law turns out to be a statement on conservation of angularmomentum, which is discussed in chapter 4. We’ll demonstratethe elliptical orbit law numerically in chapter 3, and analytically inchapter 4.

2.3.2 Circular orbits

Kepler’s laws say that planets move along elliptical paths (withcircles as a special case), which would seem to contradict the proofon page 90 that objects moving under the influence of gravity haveparabolic trajectories. Kepler was right. The parabolic path wasreally only an approximation, based on the assumption that thegravitational field is constant, and that vertical lines are all parallel.In figure e, trajectory 1 is an ellipse, but it gets chopped off whenthe cannonball hits the earth, and the small piece of it that is aboveground is nearly indistinguishable from a parabola. Our goal isto connect the previous calculation of parabolic trajectories, y =(g/2v2)x2, with Kepler’s data for planets orbiting the sun in nearlycircular orbits. Let’s start by thinking in terms of an orbit thatcircles the earth, like orbit 2 in figure e. It’s more natural nowto choose a coordinate system with its origin at the center of theearth, so the parabolic approximation becomes y = r − (g/2v2)x2,where r is the distance from the center of the earth. For smallvalues of x, i.e., when the cannonball hasn’t traveled very far fromthe muzzle of the gun, the parabola is still a good approximationto the actual circular orbit, defined by the Pythagorean theorem,r2 = x2 +y2, or y = r

√1− x2/r2. For small values of x, we can use

the approximation√

1 + ε ≈ 1+ε/2 to find y ≈ r−(1/2r)x2. Settingthis equal to the equation of the parabola, we have g/2v2 = (1/2r),or

v =√gr [condition for a circular orbit].

Low-earth orbit example 14To get a feel for what this all means, let’s calculate the velocityrequired for a satellite in a circular low-earth orbit. Real low-earth-orbit satellites are only a few hundred km up, so for purposes ofrough estimation we can take r to be the radius of the earth, andg is not much less than its value on the earth’s surface, 10 m/s2.Taking numerical data from Appendix 4, we have

v =√

gr

=√

(10 m/s2)(6.4× 103 km)

=√

(10 m/s2)(6.4× 106 m)

=√

6.4× 107 m2/s2

= 8000 m/s


(about twenty times the speed of sound).

In one second, the satellite moves 8000 m horizontally. Dur-ing this time, it drops the same distance any other object would:about 5 m. But a drop of 5 m over a horizontal distance of 8000m is just enough to keep it at the same altitude above the earth’scurved surface.

2.3.3 The sun’s gravitational field

We can now use the circular orbit condition v =√gr, combined

with Kepler’s law of periods, T ∝ r3/2 for circular orbits, to deter-mine how the sun’s gravitational field falls off with distance.8 Fromthere, it will be just a hop, skip, and a jump to get to a universaldescription of gravitational interactions.

The velocity of a planet in a circular orbit is proportional tor/T , so

r/T ∝ √grr/r3/2 ∝ √gr

g ∝ 1/r2

If gravity behaves systematically, then we can expect the same tobe true for the gravitational field created by any object, not just thesun.

There is a subtle point here, which is that so far, r has just meantthe radius of a circular orbit, but what we have come up with smellsmore like an equation that tells us the strength of the gravitationalfield made by some object (the sun) if we know how far we are fromthe object. In other words, we could reinterpret r as the distancefrom the sun.

2.3.4 Gravitational energy in general

We now want to find an equation for the gravitational energy ofany two masses that attract each other from a distance r. We assumethat r is large enough compared to the distance between the objectsso that we don’t really have to worry about whether r is measuredfrom center to center or in some other way. This would be a goodapproximation for describing the solar system, for example, sincethe sun and planets are small compared to the distances betweenthem — that’s why you see Venus (the “evening star”) with yourbare eyes as a dot, not a disk.

The equation we seek is going to give the gravitational energy,U , as a function of m1, m2, and r. We already know from expe-

8There is a hidden assumption here, which is that the sun doesn’t move.Actually the sun wobbles a little because of the planets’ gravitational interactionswith it, but the wobble is small due to the sun’s large mass, so it’s a pretty goodapproximation to assume the sun is stationary. Chapter 3 provides the tools toanalyze this sort of thing completely correctly — see p. 144.

Section 2.3 Gravitational phenomena 99

f / The gravitational energyU = −Gm1m2/r graphed as afunction of r .

rience with gravity near the earth’s surface that U is proportionalto the mass of the object that interacts with the earth gravitation-ally, so it makes sense to assume the relationship is symmetric: U ispresumably proportional to the product m1m2. We can no longerassume ∆U ∝ ∆r, as in the earth’s-surface equation ∆U = mg∆y,since we are trying to construct an equation that would be validfor all values of r, and g depends on r. We can, however, con-sider an infinitesimally small change in distance dr, for which we’llhave dU = m2g1 dr, where g1 is the gravitational field created bym1. (We could just as well have written this as dU = m1g2 dr, sincewe’re not assuming either mass is “special” or “active.”) Integratingthis equation, we have∫

dU =

∫m2g1 dr

U = m2

∫g1 dr

U ∝ m1m2

∫1

r2dr

U ∝ −m1m2

r,

where we’re free to take the constant of integration to be equal tozero, since gravitational energy is never a well-defined quantity inabsolute terms. Writing G for the constant of proportionality, wehave the following fundamental description of gravitational interac-tions:

U = −Gm1m2

r[gravitational energy of two masses

separated by a distance r]

We’ll refer to this as Newton’s law of gravity, although in realityhe stated it in an entirely different form, which turns out to bemathematically equivalent to this one.

Let’s interpret his result. First, don’t get hung up on the factthat it’s negative, since it’s only differences in gravitational energythat have physical significance. The graph in figure f could be shiftedup or down without having any physical effect. The slope of thisgraph relates to the strength of the gravitational field. For instance,suppose figure f is a graph of the gravitational energy of an asteroidinteracting with the sun. If the asteroid drops straight toward thesun, from A to B, the decrease in gravitational energy is very small,so it won’t speed up very much during that motion. Points C andD, however, are in a region where the graph’s slope is much greater.As the asteroid moves from C to D, it loses a lot of gravitationalenergy, and therefore speeds up considerably. This is due to thestronger gravitational field.


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27 - Light and Matter · moustache, in units of kg. 36 According to folklore, every time you take a breath, you are inhaling some of the atoms exhaled in Caesar’s last words. Is